An automobile battery has an emf of 12.6 V and an internal resistance of 0.0890

Question

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0890 ?. The headlights together have an equivalent resistance of 5.10 ? (assumed constant).

(a) What is the potential difference across the headlight bulbs when they are the only load on the battery? (Enter your answer to at least two decimal places.)
V

(b) What is the potential difference across the headlight bulbs when the starter motor is operated, with 35.0 A of current in the motor? (Enter your answer to at least two decimal places.)
V

Explanation / Answer

Part A.

Using Ohm's law:

V = i*Req

E = i*(R + r)

i = E/(R + r)

i = 12.6/(5.10 + 0.0890)

i = 2.428 Amp

Now Voltage drop across headlight will be

dV = i*R

dV = 2.428*5.1 = 12.38 V

Part B

In this case relation between currents will be, from kirchoff's law:

i1 = i2 + 35

Also using kirchoff's voltage law:

E = i1*r + i2*R

12.6 = (i2 + 35)*(0.0890) + i2*5.10

i2 = (12.6 - 35*0.0890)/(0.0890 + 5.10)

i2 = 1.827 Amp

Voltage drop across headlights will be

dV = i2*R = 1.827*5.10 = 9.32 V

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